Problem: Simplify; express your answer in exponential form. Assume $t\neq 0, p\neq 0$. $\dfrac{{(t^{-4})^{-5}}}{{(t^{2}p^{-1})^{-1}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{-4}}$ to the exponent ${-5}$ . Now ${-4 \times -5 = 20}$ , so ${(t^{-4})^{-5} = t^{20}}$ In the denominator, we can use the distributive property of exponents. ${(t^{2}p^{-1})^{-1} = (t^{2})^{-1}(p^{-1})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{-4})^{-5}}}{{(t^{2}p^{-1})^{-1}}} = \dfrac{{t^{20}}}{{t^{-2}p}}$ Break up the equation by variable and simplify. $\dfrac{{t^{20}}}{{t^{-2}p}} = \dfrac{{t^{20}}}{{t^{-2}}} \cdot \dfrac{{1}}{{p}} = t^{{20} - {(-2)}} \cdot p^{- {1}} = t^{22}p^{-1}$.